Chapter 6, Problem 98P

To determine

(a)

The rate of discharge through the slit.

Answer to Problem 98P

The rate of discharge through the slit is $\left(0.0204{\text{m}}^{\text{3}}/\text{s}\right)$.

Explanation of Solution

Given information:

The length of the slit is $1.2\text{m}$, width of the rectangular slit is $5\text{mm}$, water discharge velocity profile is parabolic and varying from $3\text{m}/\text{s}$ on one end to $7\text{m}/\text{s}$ in other end.

Write the expression for velocity based on parabolic profile.

  $V=k{l}^2+C$   ....... (I)

Here, the first constant is $k$, the length of the pipe is $l$ and the second constant is $C$.

Write the expression for flow rate of water.

  $d\dot{V}=dA\times V$........ (II)

Here, the elemental area is $dA$ and the velocity is $V$.

Write the expression of elemental area.

  $dA=b\times dl$   ....... (III)

Here, the width of the rectangular slit is $b$ and the elemental length of the slit is $dl$.

Write the expression of elemental flow rate of water.

  $d\dot{V}=dA\times V$   ....... (IV)

Here, the elemental area is $dA$ and the velocity is $V$.

Write the expression of the force acting on the element.

  $dF=\rho \left(dA\right)V^2$   ....... (V)

Here, the density of the fluid is $\rho$, the elemental area is $dA$ and the velocity is $V$.

Calculation:

At the starting of the jet.

Substitute $0$ for $l$ and $3\text{m}/\text{s}$ for $V$ in Equation (I).

  $\begin{array}{l}3\text{m}/\text{s}=k{\left(0\right)}^2+C\\ C+0=3\text{m}/\text{s}\\ C=3\text{m}/\text{s}\end{array}$

At the end of the jet.

Substitute $1.2\text{m}$ for $l$, $7\text{m}/\text{s}$ for $V$ and $3\text{m}/\text{s}$ for $C$ in Equation (I).

  $\begin{array}{l}7\text{m}/\text{s}=k{\left(1.2\text{m}\right)}^2+3\text{m}/\text{s}\\ k\times 1.44{\text{m}}^{\text{2}}=10\text{m}/\text{s}\\ k=2.777/\text{m}\cdot \text{s}\end{array}$

Substitute $2.777/\text{m}\cdot \text{s}$ for $k$ and $3\text{m}/\text{s}$ for $C$ in Equation (I).

  $V=\left(2.777/\text{m}\cdot \text{s}\right)l^2+3\text{m}/\text{s}$

Substitute $5\text{mm}$ for $b$ in Equation (III).

  $\begin{array}{c}dA=\left(5\text{mm}\right)\times dl\\ =\left(5\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\times dl\\ =0.005\text{m}\times dl\end{array}$

Substitute $0.005\text{m}\times dl$ for $dA$ in Equation (IV).

  $d\dot{V}=0.005\text{m}\times dl\times V$   ....... (VI)

Substitute $\left(2.777/\text{m}\cdot \text{s}\right)l^2+3\text{m}/\text{s}$ for $V$ in Equation (VI).

  $\begin{array}{c}d\dot{V}=0.005\text{m}\times dl\times \left(\left(2.777/\text{m}\cdot \text{s}\right)l^2+3\text{m}/\text{s}\right)\\ =0.005\left(2.777l^2/\text{s}+3{\text{m}}^{\text{2}}/\text{s}\right)\times dl\end{array}$   ....... (VII)

Integrate Equation (VII) on both sides the change in length is $0$ to $1.2\text{m}$ with respect to change in flow rate is $0$ to $\dot{V}$.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\dot{V}}{\int }}d\dot{V}}={\displaystyle \underset{0}{\overset{1.2\text{m}}{\int }}0.005\left(2.777l^2/\text{s}+3{\text{m}}^{\text{2}}/\text{s}\right)\times dl}\\ {\left[\dot{V}\right]}_0^{\dot{V}}=0.005{\left[\frac{l^3}{3}\right]}_0^{1.2\text{m}}\left(\text{1}/\text{s}\right)+0.015{\left[l\right]}_0^{1.2}\left({\text{m}}^{\text{2}}/\text{s}\right)\\ \dot{V}=\frac{0.005\times {\left(1.2\text{m}\right)}^3}{3}\left({\text{m}}^{\text{3}}/\text{s}\right)+0.015\times \left(1.2\right)\left({\text{m}}^{\text{3}}/\text{s}\right)\\ \dot{V}=\left(0.0204{\text{m}}^{\text{3}}/\text{s}\right)\end{array}$

Conclusion:

The rate of discharge through the slit is $\left(0.0204{\text{m}}^{\text{3}}/\text{s}\right)$.

To determine

(b)

The vertical force acting on the pipe due to discharge process.

Answer to Problem 98P

The vertical force acting on the pipe due to discharge process is $113.07\text{N}$

Explanation of Solution

Calculation:

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$

  $0.005\text{m}\times dl$ for $dA$ and $\left(2.777/\text{m}\cdot \text{s}\right)l^2+3\text{m}/\text{s}$ for $V$ in Equation (V).

  $\begin{array}{c}dF=1000\text{kg}/{\text{m}}^{\text{3}}\left(0.005\text{m}\times dl\right){\left(\left(2.777/\text{m}\cdot \text{s}\right)l^2+3\text{m}/\text{s}\right)}^2\\ =5\times dl\text{kg}/{\text{m}}^{\text{2}}\left({\left(\left(2.777/\text{m}\cdot \text{s}\right)l^2\right)}^2+{\left(3\text{m}/\text{s}\right)}^2+2\times \left(2.777/\text{m}\cdot \text{s}\right)l^2\times 3\text{m}/\text{s}\right)\\ =5\left(\left(7.71/{\text{m}}^2\cdot {\text{s}}^2\right)l^4+\left(9{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)+\left(16.62/{\text{s}}^{\text{2}}\right)l^2\right)dl\text{kg}/{\text{m}}^{\text{2}}\end{array}$   ....... (VIII)

Integrate equation (VIII) on both sides the change in length of the element is $0$ to $1.2\text{m}$ with respect to change in force is $0$ to $F$.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{F}{\int }}dF}={\displaystyle \underset{0}{\overset{1.2\text{m}}{\int }}5\left(\left(7.71/{\text{m}}^2\cdot {\text{s}}^2\right)l^4+\left(9{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)+\left(16.62/{\text{s}}^{\text{2}}\right)l^2\right)dl\text{kg}/{\text{m}}^{\text{2}}}\\ {\left[F\right]}_0^F={\displaystyle \underset{0}{\overset{1.2\text{m}}{\int }}\left(\left(38.55\times l^4/{\text{m}}^2\cdot {\text{s}}^2\right)+\left(45{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)+\left(83.1\times l^2/{\text{s}}^{\text{2}}\right)\right)}dl\text{kg}/{\text{m}}^{\text{2}}\\ F=\frac{38.55}{5}{\left[l^5\right]}_0^{1.2\text{m}}\left(\text{kg}/{\text{m}}^4\cdot {\text{s}}^2\right)+\left(45\right){\left[l\right]}_0^{1.2\text{m}}\left(\text{kg}/{\text{s}}^{\text{2}}\right)+\frac{83.1}{3}{\left[l\right]}_0^{1.2\text{m}}\left(\text{kg}/{\text{m}}^{\text{2}}\cdot {\text{s}}^{\text{2}}\right)\\ F=7.71\times {\left(1.2\text{m}\right)}^5\left(\text{kg}/{\text{m}}^4\cdot {\text{s}}^2\right)+45\times \left(1.2\text{m}\right)\left(\text{kg}/{\text{s}}^{\text{2}}\right)+27.7\times {\left(1.2\text{m}\right)}^2\left(\text{kg}/{\text{m}}^{\text{2}}\cdot {\text{s}}^{\text{2}}\right)\end{array}$

  $\begin{array}{c}F=7.71\times {\left(1.2\text{m}\right)}^5\left(\text{kg}/{\text{m}}^4\cdot {\text{s}}^2\right)+45\times \left(1.2\text{m}\right)\left(\text{kg}/{\text{s}}^{\text{2}}\right)+27.7\times {\left(1.2\text{m}\right)}^2\left(\text{kg}/{\text{m}}^{\text{2}}\cdot {\text{s}}^{\text{2}}\right)\\ =\left(19.18\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)+\left(54\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)+\left(39.89\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(113.07\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =113.07\text{N}\end{array}$

Conclusion:

The vertical force acting on the pipe due to discharge process is $113.07\text{N}$